The wheel on an upside-down bicycle moves through 18.0 rad in 5.39 s. What is the wheel’s angular acceleration if its initial angular speed is 2.5 rad/s? Answer in units of rad/s 2 .

Answer :

Explanation:

Formula to calculate angular acceleration is as follows.

   [tex]\Delta (\theta) = \frac{1}{2} \alpha \Delta t^{2} + \omega_{1} \Delta t[/tex]

or,      [tex]\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}[/tex]

Putting the given values into the above formula as follows.

    [tex]\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}[/tex]

                = [tex]\frac{2(18.0 rad - 2.5 rad/s \times 5.30 s)}{(5.39)^{2}}[/tex]    

                = 0.326 [tex]rad/s^{2}[/tex]

Thus, we can conclude that the wheel’s angular acceleration if its initial angular speed is 2.5 rad/s is 0.326 [tex]rad/s^{2}[/tex].

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